魔方难题（定期发，纯原创）

Atato

我觉得这个不是数学问题...

bardy

Cube subgroups

Subgroups generated by single moves

Suppose not all possible cube moves are allowed, for example suppose that you only use moves of the Up and Right faces. You can mix a cube that way, and then try to solve it again using only those moves. Clearly not all cube positions can be reached, since the 2×3×3 block at the bottom left is never disturbed by the U and R moves. In the same way we might restrict ourselves to using only half turns. There are many choices as to which faces we allow, and whether each face uses only half turns or not. Below is a list containing all possibilities.
        Generators        Size                Corners                Edges                Restrictions
0.        -        1        =        1        *        1               
1a.        U2        2        =        2        *        2        /        2
1b.        U        4        =        4        *        4        /        4
2a.        U2,R2        12        =        3!        *        3! 22/2        /        3!
2b.        U,R2        14400        =        6!/6        *        5! 2!        /        2
2c.        U,R        73483200        =        6!/6 36/3        *        7!        /        2
3a.        U2, D2, F2        96        =        4!        *        23        /        2
3b.        U2, R2, F2        2592        =        4!        *        3! 3! 3!        /        2
3c.        U, R2, F2        10886400        =        7!        *        6! 3!        /        2
3d.        U, R2, L2        58060800        =        8!        *        6! 2! 2!        /        2
3e.        U2, R, L2        58060800        =        8!        *        6! 2! 2!        /        2
3f.        U, R2, D        1625702400        =        8!        *        8! 2!        /        2
3g.        U2, R, F        666639590400        =        7! 37/3        *        9!        /        2
3h.        U, R, D2        3555411148800        =        8! 38/3        *        8! 2!        /        2
3i.        U, R, D        159993501696000        =        8! 38/3        *        10!        /        2
3j.        U, R, F        170659735142400        =        7! 37/3        *        9! 29/2        /        2
4a.        U2, D2, F2, B2        192        =        4!        *        4! 24/2        /        4!
4b.        U2, D2, F2, R2        165888        =        4! 4        *        4! 4! 3!        /        2
4c.        U, R2, L2, D2        116121600        =        8!        *        6! 2! 2! 2!        /        2
4d.        U2, D2, F2, R        2438553600        =        8!        *        7! 4!        /        2
4e.        U, R2, L2, D        3251404800        =        8!        *        8! 2! 2!        /        2
4f.        U, D2, F2, R2        4877107200        =        8!        *        8! 3!        /        2
4g.        U, D, F2, R2        4877107200        =        8!        *        8! 3!        /        2
4h.        U, D2, F2, R        1759928518656000        =        8! 38/3        *        11!        /        2
4i.        U, D, F2, R        1759928518656000        =        8! 38/3        *        11!        /        2
4j.        U2, D2, F, R        1759928518656000        =        8! 38/3        *        11!        /        2
4k.        U2, D, F, R        1802166803103744000        =        8! 38/3        *        11! 211/2        /        2
4l.        U, D, F, R        1802166803103744000        =        8! 38/3        *        11! 211/2        /        2
5a.        U2, F2, B2, R2, L2        663552        =        4! 4        *        4! 4! 4!        /        2
5b.        U, D2, F2, B2, R2        19508428800        =        8!        *        8! 4!        /        2
5c.        U, D, F2, B2, R2        19508428800        =        8!        *        8! 4!        /        2
5d.        U, F2, B2, R2, L2        19508428800        =        8!        *        8! 4!        /        2
5e.        U2, F, B2, R, L2        21119142223872000        =        8! 38/3        *        12!        /        2
5f.        U2, F, B2, R, L        21119142223872000        =        8! 38/3        *        12!        /        2
5g.        U2, F, B, R, L        21119142223872000        =        8! 38/3        *        12!        /        2
5h.        U2, D2, F, B, R        21119142223872000        =        8! 38/3        *        12!        /        2
5i.        U2, D2, F, B2, R        21119142223872000        =        8! 38/3        *        12!        /        2
5j.        U, F, B2, R, L2        43252003274489856000        =        8! 38/3        *        12! 212/2        /        2
5k.        U, F, B, R, L2        43252003274489856000        =        8! 38/3        *        12! 212/2        /        2
5l.        U, F, B, R, L        43252003274489856000        =        8! 38/3        *        12! 212/2        /        2
6a.        U2, D2, F2, B2, R2, L2        663552        =        4! 4        *        4! 4! 4! / 2        /        1
6b.        U, D, F2, B2, R2, L2        19508428800        =        8!        *        8! 4!        /        2
6c.        U, D2, F2, B2, R2, L2        19508428800        =        8!        *        8! 4!        /        2
6d.        U2, D2, F, B, R, L        21119142223872000        =        8! 38/3        *        12!        /        2
6e.        U2, D2, F, B2, R, L        21119142223872000        =        8! 38/3        *        12!        /        2
6f.        U2, D2, F, B2, R, L2        21119142223872000        =        8! 38/3        *        12!        /        2
6g.        U, D2, F, B2, R, L2        43252003274489856000        =        8! 38/3        *        12! 212/2        /        2
6h.        U, D2, F, B, R, L2        43252003274489856000        =        8! 38/3        *        12! 212/2        /        2
6i.        U, D2, F, B, R, L        43252003274489856000        =        8! 38/3        *        12! 212/2        /        2
6j.        U, D, F, B, R, L        43252003274489856000        =        8! 38/3        *        12! 212/2        /        2

The list above is complete up to turns of the whole cube, e.g. <F,R> is not included because it is isomorphic to the group generated by U and R, which is on the list. I have also omitted the groups generated by R and L, R2 and L, or R2 and L2, because those trivially decompose.

The number of reachable positions (the size of the group) is shown, as well as the number of ways the corners or the edges can be arranged separately. The last column is a factor showing how much the edges and corners restrict each other. Usually this is a factor 2, due to the normal parity restriction that the parity of the total corner and edge permutation must be even. Thus once the corners have been solved the parity of the edges is forced, and they have half as many possible arrangements than they had before the corners were solved. In some rare cases the edges and corners influence each other by more than this.

Some of the groups listed are actually identical. This can easily be shown using the following two move sequences.
1. D = F2R2D2F2U2R2F2 U F2R2U2F2D2R2F2 (uses U, F2, R2, D2)
2. D2 = F2R2L2B2U2F2R2L2B2 (uses F2, B2, R2, L2, U2)
For example 4f=<U, D2, F2, R2> and 4g=<U, D, F2, R2> are the same because with the first move sequence you can use only U, D2, F2, and R2 to get the same effect as D (and conversely with D you can of course get D2). Similarly 4h, 4i are equal. In fact, using these two move sequences you can show that any two groups on the list of the same size are identical groups, except that 4j is slightly different from 4h/4i. In group 4j the edge orientations differ from those in 4h/4i, though in all other respects they are the same. Thus they are isomorphic groups.

If you wish to solve a position using the same set of moves you used to mix them, it is often easiest to solve the corners first, and then the edges. Below is a table of useful move sequences. With conjugates of these, all positions in the above groups can be solved. Note that the sequences for the corners may disturb edges.
Effect         Generators        Sequence
URF- URB+         U, R         RU'RU'RU' R'UR'UR'U
(UFL, DFR, DBR)         U, R2, F2         UF2U'R2UF2U'F2
(URF, DRB, DBL)         U, R2, L2         UL2U'R2 UL2U'R2
(URF, DRB, DBL)         U2, R, L2         R2U2R'L2U2R'L2U2RL2U2RL2U2
UF+ UR+         U, R, F         RU'R2UFRUF2U'FR2F2R2F2
(UF, UR, UB)         U, R         RU2RURUR2U'R'U'R2
(UF, UR, UB)         U, R2         U2R2U2R2 UR2UR2 U2R2U2R2 UR2UR2
(UF, UB)(RF, RB)        U2, R2         R2U2 R2U2 R2U2
(FL, FR, BR)         U2, R2, F2        F2U2R2U2 F2U2R2U2

Subgroups generated by slice/anti-slice moves

The groups generated by slice moves, or anti-slice moves are interesting because there are many nice patterns in these groups.
        Generators        Size                Corners                Edges                Restrictions
a.        U2D2, F2B2, R2L2         8        =        4        *        23        /        4
b.        UD', FB', RL'         768        =        24        *        83/2 3!/2        /        24
c.        UD, FB, RL         6144        =        4·24        *        43 3! 23/2        /        4!
d.        UD, FB, RL, UD', FB', RL'        15925248        =        3! 4! 4        *        4! 4! 4! 3! 23/2        /        3!

The first of these, the slice-squared group, is trivial. The second, the slice group, is easy to understand if you consider the corners as fixed in space and the centres as moving pieces. Just line up each slice of edges with the corners, and you will be left with a spot pattern.

The third on the list is the anti-slice group. Positions in this group are more difficult to solve. First solve the corners relative to each other, which is in essence similar to solving the corners in the square group 6a. Next orient the edges using the move sequence RL UD FB RL UD FB, which flips the edges in the U/D layers. You should now have a cube with only opposing colours on each face. Line up the centres with the corners using squared (anti-)slices. The edges can be solved using the sequence RL UD F2B2 UD RL which is a 4H pattern on the sides.

The last group on the list is generated by all slice and anti-slice moves. Note that combining a slice and an anti-slice move you get a half turn of a single face, e.g. RL RL' = R2, so the square group (6a) is a subgroup of this one. It is relatively easy to solve by bringing it to a position in the square group. The edge flip in the previous paragraph is useful for this.

Various small subgroups

For many small finite groups, an isomorphic group can be found on the cube. For example a cyclic group of order 3 can be found by a 3-cycle of pieces. Below I will list some small groups and give generators on the cube that give such a group.
Group        Generator
C2        Any half turn such as R2, or any edge flip.
C3        Any 3-cycle such as R2UD'F2U'D, or any corner twist.
C4        Any face turn such as R
C2×C2        R2L2 and F2B2
C5        Any 5-cycle, for example D2R2D'R2L2UR'L'B2RL'.
C6=C2×C3        R2F2
S3        R2F2R2F2 and R2
C7        Any 7-cycle
C8        Any 8-cycle (combined with a 2-cycle for parity reasons), e.g. R2F2 D2R2D2 B2R2 D R2L2 U F2B2, or flipped edge-4-cycle, e.g. L2D R2F2B2 L'R' F'L2F LR' F2B2U'.
C2×C4        R2, L
C2×C2×C2        R2L2, F2B2, U2D2.
D4        UD', and FB R2L2 F'B'
Q, Quaternion group        i = (UR,UF)+(UL,UB)+ = R2 UR'U'R F R2 URU' F' UR2, j = (UB,UF)+(UR,UL)+ = L'U'B' U2 BLU FU2F', k = (UL,UF)+(UB,UR)+ = B'U'R U2 FRF'R' U' BRU'R'. Note that they all have order 4 and that ijk=jki=kij=ii=jj=kk.
A4        R2F2B2R2F2B2 and F2R2F2R2 gives all even permutations of the 4 corner columns.
C13        Impossible. 13 does not divide the order of the cube group.
C2×C2×C4        U2, D, Superflip.
C2×C2×C2×C2        U2D2, F2B2, R2L2, Superflip
D4×C2        UD', FB R2L2 F'B', and R2F2B2L2F2B2
C213        The 6H pattern F2R2L2B2L2R2 in its 6 possible orientations, the double corner swap U R2FR2 U2D2 L2BL2 UD2 in its 6 possible orientations, and 4-flip UF2D F2B2 DL2D F'B LB2L2 U'D F'R'L in its three possible orientations.

bardy

这个列表应该比较详细了，具体是否正确需要自己判断了

lvyingru

搞不懂。。  

无为子

高手大把，俺只来看看，正在学习中···ing

noski

楼主的问题，U2、D2、F2、B2、R2、L2，这六个操作都使用一次，不同的组合，共有42个状态，程序求得，计算公式我给不出来。

另外，12楼的表格不错！
最少步区讨论的1x3x3魔方，正是三阶魔方的（U2,D2,F2,B2）子状态集，有192个状态。

---------------------------------
更正，这42的结果是不对的，只算了角块，没有算棱块。

[ 本帖最后由 noski 于 2009-7-24 18:23 编辑 ]

migl

U2 D2 F2 B2 L2 R2
这六个“数”进行排列组合，然后减去相同的状态数即可。
（ 我也感觉是不多于：6！=720，但又要少很多。 ）

比如：
U2 D2 F2 B2 L2 R2
D2 U2 F2 B2 L2 R2
U2 D2 F2 B2 L2 R2
U2 D2 B2 F2 L2 R2
U2 D2 F2 B2 L2 R2
U2 D2 F2 B2 R2 L2
这六种是同一个状态。（ 即：对称棋盘 ）
而：
U2 D2 F2 B2 L2 R2
U2 D2 L2 R2 F2 B2
F2 B2 U2 D2 L2 R2
F2 B2 L2 R2 U2 D2
L2 R2 U2 D2 F2 B2
L2 R2 F2 B2 U2 D2
都是“对称棋盘”。
所以至少有6*6=36条公式是“对称棋盘”，是同一种状态。

但是， L2 F2 R2 B2 U2 D2 却不是“对称棋盘”。

唉，本人脑细胞有限。等高手们继续发挥吧~~~